Acetone on treatment with `CH_(3)-Mg-I` and on further hydrolysis gives

Options

A. Isopropyl alcohol
B. Primary alcohol
C. Acetic acid
D. 2 - methyl 2 - propanol

Answer

To solve the question, we need to understand the reaction of acetone with methyl magnesium iodide (a Grignard reagent) and the subsequent hydrolysis step. Here’s a step-by-step breakdown of the process:

### Step 1: Identify the Reactants
- The reactant is acetone, which is chemically represented as \( CH_3COCH_3 \).
- The Grignard reagent used is methyl magnesium iodide, represented as \( CH_3MgI \).

### Step 2: Understand the Mechanism of Reaction
- Grignard reagents act as nucleophiles. In this case, the carbon atom attached to the magnesium in \( CH_3MgI \) is nucleophilic due to the polarization of the bond between carbon and magnesium.
- Acetone has a carbonyl group (C=O), where the carbon atom is electrophilic due to the electronegativity of oxygen.

### Step 3: Nucleophilic Addition
- The nucleophilic carbon from \( CH_3MgI \) attacks the electrophilic carbon of the carbonyl group in acetone.
- This results in the formation of a tetrahedral intermediate:
\[ CH_3C(OH)(CH_3)MgI \]
- Here, the oxygen atom becomes negatively charged due to the addition of the nucleophile.

### Step 4: Hydrolysis
- The next step involves hydrolysis of the intermediate. When the intermediate is treated with water (or an acid like \( H_3O^+ \)), the negatively charged oxygen will pick up a proton (H⁺) from water.
- This results in the formation of the alcohol:
\[ CH_3C(OH)(CH_3) \rightarrow 2-methyl-2-propanol \]

### Step 5: Identify the Product
- The final product after hydrolysis is 2-methyl-2-propanol, which can also be represented as \( (CH_3)_2C(OH)CH_3 \).

### Conclusion
- Therefore, the product formed when acetone reacts with methyl magnesium iodide and undergoes hydrolysis is **2-methyl-2-propanol**.

### Final Answer
The correct option is **4. 2-methyl-2-propanol**.

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